$f(x) = \int_{0}^{+\infty} \frac{x(s)}{\cosh(s)} ds$ is bounded for $x \in L_2(0 , + \infty)$. And find its norm. The question is as follows:
Prove the linear boundedness of $f(x) = \int_{0}^{+\infty} \frac{x(s)}{\cosh(s)} ds$, for $x \in L_2(0 , + \infty)$. And find its norm.
$\textbf{some effort:}$
For to show it is bounded, we have $||fx||^2 = \int_{0}^{+\infty} \mid \int_{0}^{+\infty} \frac{x(s)}{\cosh(s)} ds \mid^2 dt$
$\hspace{7.6cm} \leq \int_{0}^{+\infty} \int_{0}^{+\infty} \mid \frac{x(s)}{\cosh(s)} \mid^2 ds dt$
$\hspace{7.6cm} = \int_{0}^{+\infty} \int_{0}^{+\infty} \frac{1}{\cosh(s)} \mid x(s) \mid^2 ds dt$
$\hspace{7.6cm} = \int_{0}^{+\infty} \frac{1}{\cosh(s)} \mid x(s) \mid^2 \int_{0}^{s} 1  dt ds $
$\hspace{7.6cm} = \int_{0}^{+\infty} \frac{s}{\cosh(s)} \mid x(s) \mid^2 ds $
 $\hspace{7.6cm} \leq  \bigg(\int_{0}^{+\infty} \frac{s^2}{\cosh^2(s)} ds\bigg)^{\frac{1}{2}} \bigg(\int_{0}^{+\infty} \mid x(s) \mid^2 ds \bigg)^{\frac{1}{2}} $ 
Which will imply that $||f|| \leq \bigg(\int_{0}^{+\infty} \frac{s^2}{\cosh^2(s)} ds\bigg)^{\frac{1}{4}}  $.
If I am right until now, then we need to calculate $\int_{0}^{+\infty} \frac{s^2}{\cosh^2(s)} ds$.
Can you please let me know if my calculation is far behind being correct?
And can you please let me know that how can I find its norm?
Thanks!
 A: Ok, it seems that you are misunderstanding what you want to do.
If $f$ is defined as you did then it is a linear operator $f:L^2(0,+\infty)\rightarrow \mathbb R$, so when you fix $x^* \in L^2(0,+\infty) \rightarrow f(x^*)=\int_{0}^{+\infty} \frac{x^*(s)}{cosh(s)}ds \in \mathbb R$! So it makes no sense write $\|fx\|^2=\int_{0}^{+\infty}|\int_{0}^{+\infty}\frac{x(s)}{cosh(s)}ds|^2dt$, in fact $f(x) \notin L^2$. 
The natural norm of $f(x)$ is the natural norm in $\mathbb R$ which is the absolute value, now we want to show that:
$$|f(x)|\le k \cdot \|x\|_{L^2(0,+\infty)}, \forall x \in L^2(0,+\infty).$$
Now notice that the function $y(t)=\frac{1}{cosh(t)} \in L^2(0,+\infty)$ so thanks to cauchy schwarz's inequality we have:
$$|f(x)|=<x,y>_{L^2(0,+\infty)}\le \|x\|_{L^2(0,+\infty)} \|y\|_{L^2(0,+\infty)}$$
so $f $ is bounded, now take $x=y$:
$$|f(y)|=<y,y>=\|y\|^2_{L^2(0,+\infty)}=\|y\|_{L^2(0,+\infty)}\|y\|_{L^2(0,+\infty)}.$$
So we also have that $\|f\|=\|y\|_{L^2(0,+\infty)}$.
Leave aside the exercise, it's very important that you understand the first part because otherwise you won't be able to solve these kind of problems.
